You have one chocolate bar that consists of some chunks. Each chunk has its own sweetness given by the array sweetness.
You want to share the chocolate with your k friends so you start cutting the chocolate bar into k + 1 pieces using k cuts, each piece consists of some consecutive chunks.
Being generous, you will eat the piece with the minimum total sweetness and give the other pieces to your friends.
Find the maximum total sweetness of the piece you can get by cutting the chocolate bar optimally.
Solution
将巧克力分为 /(k+1/) 份，其中你得到的是最小的那部分，问最大可能的这部分是多少。
我们二分答案：每次对当前的答案进行 /(check/)，也就是看是否能够分成 /(k+1/) 份
点击查看代码
class Solution {
private:
bool check(vector<int>& sw, int cand, int k){
int sum=0;
for(auto ele:sw){
sum+=ele;
if(sum>=cand){
k--;
if(k==0)return true;
sum=0;
}
}
return false;
}
public:
int maximizeSweetness(vector<int>& sweetness, int k) {
k++;
int n = sweetness.size();
int l = 0,r=0;
for(int i=0;i<n;i++)r+=sweetness[i];
int ans=0;
while(l<=r){
int mid = (r-l)/2+l;
if(check(sweetness, mid, k)){
ans=mid;
l=mid+1;
}
else r=mid-1;
}
return ans;
}
};
原创文章，作者：ItWorker，如若转载，请注明出处：https://blog.ytso.com/289979.html